13

Let us agree to say that $\mu$ is a Radon measure on a metric space $X$ if it is a Borel measure which is finite on compact subsets and is such that:

  • Every measurable subset $A$ is outer regular, meaning that $$\mu(A)=\inf\{\mu(V)\ |\ A\subset V,\ V\ \text{is open}\};$$
  • Every open subset $U$ is inner regular, meaning that $$\mu(U)=\sup\{\mu(K)\ |\ K\subset U,\ K\ \text{is compact}\}.$$

The present question regards inner regularity. Indeed, as I read in Evans-Gariepy's Measure theory and fine properties of functions, Theorem 4, Chapter 1 (*), if $X=\mathbb{R}^n$ then every measurable subset is automatically inner regular. Is this a property of $\mathbb{R}^n$ alone? Formally:

Question. Which metric spaces have the property that for any Radon measure every measurable subset is inner regular?

(*) Notation and conventions in this book are a bit different from the ones of the present post.

EDIT. Specifically, in Evans-Gariepy's book a measure is a extended-real valued set function which is monotone and subadditive (usually, this is called a outer measure). A measurable set is one which satisfies Caratheodory's criterion: $$E\ \text{is measurable} \iff \forall T\subset X,\ \mu(T)=\mu(T\cap E)+ \mu(T\cap E^c).$$ A Radon measure is a (outer) measure which is:

  1. Borel regular, meaning that every Borel set is measurable and every set (even nonmeasurable ones) is contained in a Borel set of the same (outer) measure;
  2. Finite on compact subsets.

The aforementioned Theorem 4 of Chapter 1 says that, given a Radon (outer) measure on $\mathbb{R}^n$, every set (measurable or not) is outer regular and every measurable set is inner regular.

Remark 1. If $\mu$ is a Borel regular measure on $X$ such that every Borel set is inner regular, then every measurable subset of finite measure is inner regular. Indeed, if $M\subset X$ is measurable and has finite measure, then by applying two times the Borel regularity property we can get a Borel set $M'$ which is contained in $M$ and has the same measure as $M$.

Remark 2. In particular, if $X$ is locally compact and is expanding union of compact sets, as in Micheal's kind answer below, and if $\mu$ is a Borel regular Radon (outer) measure, I believe that every (Caratheodory) measurable subset is inner regular. Indeed let $M\subset X$ be measurable. If $\mu(M)<\infty$ we are done. If $\mu(M)=\infty$ then we can write it as an expanding union of sets of finite measure: $M=\cup_1^\infty M_j$. Every $M_j$ contains a compact $K_j$ such that $\mu(K_j)\ge \mu(M_j)-1$. Letting $j\to \infty$, $\mu(M_j)\to \mu(M)=\infty$ and so $\mu(K_j)\to \infty$ too. This proves the claim.

Giuseppe Negro
  • 32,319
  • What kind of spaces are you interested in? Do you care about large spaces (not $\sigma$-compact)? Would you mind adding completeness? Have you considered the example of Lebesgue measure times counting measure on the reals times the discrete reals (it has two incarnations: one is inner regular but not outer regular and the other is outer regular but not inner regular)? – t.b. Dec 19 '12 at 08:00
  • I didn't work it out in full detail, but I think that one can use the example I alluded to above to show that on every non-separable complete metric space without isolated points there is an outer measure as in Evans-Gariepy, but that the associated measure (from Carathéodory) will not be tight while satisfying all your conditions. On the other hand, one can always construct a tight version of it which will fail to be outer regular. – t.b. Dec 19 '12 at 10:27
  • @t.b.: Actually I am exploring some measure theory because of a course in fractal geometry which I am attending. So the spaces I am most interested in usually are $\sigma$-compact. For those spaces, if I understand well Micheal's answer below, all Radon measures are automatically tight (meaning that any measurable set is inner regular). By all means it would be nice to construct your counterexample so as to put a period to this question. I'll think about it. Thank you! – Giuseppe Negro Dec 20 '12 at 02:58
  • I see. Here's the idea: The example of a non-tight Radon measure on an uncountable disjoint union of $\mathbb{R}$ I linked you to can be modified to use the Cantor set with the "coin-flipping measure" instead of $\mathbb{R}$. Thus the task is essentially reduced to find a closed subspace of a metric space which is homeomorphic to an uncountable disjoint union of Cantor sets. This can be done using this construction and a locally finite disjoint family of open sets (and some effort). I'm not so sure if the actual construction is all that enlightening. – t.b. Dec 20 '12 at 04:31
  • You understand Michael's answer correctly. You can find proofs in many places, for example Theorem 3.2 in Parthasarathy's Probability Measures on Metric Spaces or Theorem 17.11 of Kechris's Classical Descriptive Set Theory. – t.b. Dec 20 '12 at 04:39
  • This concept of Radon topological space is exactly the same thing this question is about: https://en.wikipedia.org/wiki/Radon_space – Giuseppe Negro Feb 25 '16 at 10:24

2 Answers2

5

Every finite measure on a topological space is outer regular if and only if it is inner regular, so compactness would be sufficient (and every finite measure on metric space is automatically both inner and outer regular). Another sufficient condition is that the space is the union of an increasing sequence of compact sets, which covers the case of $\mathbb{R}^n$.

One can look this up in: Aliprantis & Border 2006, "Infinite Dimensional Analysis" in section 12.1.

Edit: Lurker has pointed out that aliprantis and Border use the weaker notion of inner regularity of apprximating sets from below by closed sets. What is called inner regularity here, they call tightness. It is automatic for finite measures on Polish spaces but not all metric spaces.

Michael Greinecker
  • 32,841
  • 6
  • 80
  • 137
  • 1
    A finite measure on metric space is both inner and outer regular for Borel sets, but does this extend to all measurable sets? If yes, how would you do it? – T. Eskin Dec 14 '12 at 09:43
  • 1
    @ThomasE: The question was about Borel sets. Inner regularity is straightforward since every measurable set can be written as the union of a Borel set and a null-set. If a set can be approximated from within by compact sets, it can (in a Hausdorff space) also be approximated from within by closed sets. So applying the argument to the complement of a measurable set gives you outer regulariy. – Michael Greinecker Dec 14 '12 at 09:57
  • Note that Aliprantis & Border define Borel measure differently than Evans-Gariepy. In the first, a Borel measure is a measure such that the measurable sets are exactly the Borel sets, and in the latter, Borel measure is an outer-measure so that the collection of measurable sets contains Borel sets. So the question was about a broader collection than just Borel sets. And thanks for providing an argument, I will try to follow it in a second :-) – T. Eskin Dec 14 '12 at 10:08
  • @ThomasE. Giuseppe wrote in the first sentence "Let us agree to say that $\mu$ is a Radon measure on a metric space $X$ if it is a Borel measure which[...]" and the last sentence is "Notation and conventions in this book are a bit different from the ones of the present post." So I did not focus on the definition in the book. – Michael Greinecker Dec 14 '12 at 10:10
  • Could you clarify a small point in your argument. You said that every measurable set can be written as the union of a Borel set and a null-set. But the following puzzles me, if you find it relevant. Let $A\subset [0,1]$ be a non-Lebesgue measurable and define an outer-measure on $\mathbb{R}$ by $\tilde{\nu}=\mu|_{A}$, where $\mu$ is the Lebesgue outer−measure. Let $\nu$ be the measure obtained from $\tilde{\nu}$. It is a good exercise to show that $\nu$ is a Radon measure but it is not Borel−regular, i.e. every measurable set $B$ does not have a Borel set $F\supset B$ with $\nu(B)=\nu(F)$. – T. Eskin Dec 14 '12 at 11:00
  • In particular the above measure is a finite Borel measure on a $\sigma$-compact metric space. – T. Eskin Dec 14 '12 at 11:07
  • @Thomas The statement that every measurable set is the union of a Borel set and a null set refers to completions of measures defined on the Borel sets. So basically, the argument says that regularity on the Borel sets implies regularity of the completion. I don't know how to relate that to what Evans-Gariepy do. – Michael Greinecker Dec 14 '12 at 11:40
  • I am sorry about this clash of notations. Indeed I had Evans-Gariepy's definition in the back of my head, but since this is less standard (AFAIK) I preferred to focus on Borel sets only. I'm adding a remark in the main text to clarify this a little. – Giuseppe Negro Dec 14 '12 at 12:20
  • 1
    If you regard a Borel measure to be only on the Borel sets, as the OP has now clarified, then I can see your point. But in general when Borel measures are defined on a broader $\sigma$-algebra that contains the Borel sets, then Radon and Borel-regularity are two properties that neither imply the other. In that case I don't see immediately how completions are related, but I will do my best to look into it. Thanks for providing the clarification. – T. Eskin Dec 14 '12 at 12:24
  • @ThomasE.: I have added a small post-scriptum to the main post. – Giuseppe Negro Dec 14 '12 at 12:47
  • This should be a comment to Michael Greinecker's answer: Aliprantis and Border's notion of inner regularity is inner regularity with respect to closed sets, as opposed to inner regularity with respect to compact sets. With this definition the equivalence you mention is a complete triviality, i.e., a matter of taking complements. –  Dec 15 '12 at 00:14
  • It is far from true that outer regularity implies tightness (inner regularity with respect to compact sets). The standard counterexample is to take a Lebesgue non-measurable set $E$ of zero inner measure in $[0,1]$ and the restriction of Lebesgue measure to $E$. This yields an outer regular Borel measure which is atomic, i.e., every Borel set has measure $0$ or $1$. This measure is outer regular because Lebesgue measure has that property, but it is not tight since every compact subset of $E$ has measure zero. –  Dec 15 '12 at 00:14
  • @lurker Thak you! I stand corrected. – Michael Greinecker Dec 15 '12 at 10:19
  • @lurker: Can you please clarify what you mean by restriction? The definition I know, again taken from Evans-Gariepy's book, is the following: $$\mathcal{L}^1\lfloor_E(B)=\mathcal{L}^1(E\cap B), \qquad \forall B\subset \mathbb{R}, $$ where $\mathcal{L}^1$ is the Lebesgue (outer) measure on the line. Do you mean the same thing? – Giuseppe Negro Dec 15 '12 at 18:45
  • What goes wrong when using closed sets instead of compact ones in the def of inner measure? – Vivaan Daga Jun 21 '23 at 21:37
3

In every locally compact $\sigma$-compact space a Radon measure, as you defined it, is inner regular (on all - not just open - measurable sets). This is Corollary 7.6 from Real Analysis by Folland (who uses the same definitions as you). You don't need it to be a metric space.

(Although it's nice to note that if it is a metric space, in addition to being locally compact $\sigma$-compact, then all locally finite Borel measures satisfy these properties by Corollary 7.8 of the same book).

Note: this might be exactly what Greinecker said, but it seems he referred to a book using a different definition, so I thought this might be helpful.

Cronus
  • 3,336